The minimum value of


The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha$ has at least one solution in $\left(0, \frac{\pi}{2}\right)$ is



$f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$

Let $\sin x=t \quad \because x \in\left(0, \frac{\pi}{2}\right) \Rightarrow 0





$=\frac{(3 t-2)(2-t)}{t^{2}(1-t)^{2}}$

$f_{\min }$ at $t=\frac{2}{3}$

$\alpha_{\min }=f\left(\frac{2}{3}\right)=\frac{4}{\frac{2}{3}}+\frac{1}{1-\frac{2}{3}}$





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